Whats up mates , i have already started to learn SQL database thing and i am confused here . i have to create a table with number of incidents per month.
I already know how to create table but the rest ?
SELECT
EXTRACT(month FROM dateofcall) AS x,
incidentnumber,
dateofcall
FROM
incidents
GROUP BY
incidentnumber,
x
ORDER BY
x ASC;
But its not giving me the results of incidents number per month . =(
It looks like you are grouping by too many items in your GROUP BY clause, and you are not COUNTing your incidents, just showing their details.
Try this:
SELECT EXTRACT(month FROM dateofcall) AS x,
COUNT(*) AS incidents
FROM
incidents
GROUP BY
EXTRACT(month FROM dateofcall)
ORDER BY
EXTRACT(month FROM dateofcall)
SELECT
EXTRACT(month FROM dateofcall) AS theMonth,
COUNT(*) AS theNumberOfIncidents
FROM
incidents
GROUP BY
EXTRACT(month FROM dateofcall)
ORDER BY
theMonth
Your original query wasn't counting anything. You were also grouping by incidentNumber which I assume is your primary-key, which is a nonsensical operation.
Due to a quirk in the SQL language you cannot use a column alias in GROUP BY statements, which is why you need to duplicate the EXTRACT(month FROM dateofcall) code.
Related
I need to group different years in my dataset so that I can see the total number of login_log_id each year has(BigQuery)
SELECT login_log_id,
DATE(login_time) as login_date,
EXTRACT(YEAR FROM login_time) as login_year,
TIME(login_time) as login_time,
FROM `steel-time-347714.flex.logs`
GROUP BY login_log_id
I want to make a group by so that I can see total number of login_log_id generated in different years.
My columns are login_log_id, login_time
I am getting following error :-
SELECT list expression references column login_time which is neither grouped nor aggregated at [2:6]
The error is because every column you refer to in the select need to be aggregated or be in the GROUP BY.
If you want the total logins by year, you can do:
SELECT
EXTRACT(YEAR FROM login_time) as login_year,
COUNT(1) as total_logins,
COUNT(DISTINCT login_log_id) as total_unique_logins
FROM `steel-time-347714.flex.logs`
GROUP BY login_year
But if you want the total by login_log_id and year:
SELECT
login_log_id,
EXTRACT(YEAR FROM login_time) as login_year,
COUNT(1) as total_logins
FROM `steel-time-347714.flex.logs`
GROUP BY login_log_id, login_year
I have a table called "sales" with two columns: transaction_date, and transaction_amount: VALUES ('2020-01-16 00:05:54.000000', '122.02'), ('2020-01-07 20:53:04.000000', '1240.00')
I want to find the 3-day moving average for each day in January 2020. I am returning the error that transaction_amount must be included in either an aggregated function or in the group by. It does not make sense to group by it, as I only want one entry per day in the resulting table. In my code, I already have the amount in the aggregate function SUM, so I am not sure what else to try. Here is my query so far:
SELECT EXTRACT(DAY FROM transaction_time) AS Jan20_day, SUM(transaction_amount), SUM(transaction_amount) OVER(ORDER BY EXTRACT(DAY FROM transaction_time) ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) AS rolling_average FROM sales WHERE EXTRACT(MONTH FROM transaction_time)=1 AND EXTRACT(YEAR FROM transaction_time)=2020 GROUP BY EXTRACT(DAY FROM transaction_time)
Any insight on why I am returning the following error?
Query Error: error: column "transactions.transaction_amount" must appear in the GROUP BY clause or be used in an aggregate function
I would expect something like this:
SELECT EXTRACT(DAY FROM transaction_time) AS Jan20_day,
SUM(transaction_amount),
SUM(SUM(transaction_amount)) OVER (ORDER BY EXTRACT(DAY FROM transaction_time) ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) AS rolling_average
FROM sales
WHERE transaction_time >= DATE '2020-01-01' AND
transaction_time < DATE '2020-02-01'
GROUP BY EXTRACT(DAY FROM transaction_time);
But the basic issue with your query is that you need to apply the window function to SUM(), so SUM(SUM(transaction_amount)) . . . .
There is need to use GroupBy Before Where Clause
GROUP BY clause is used with the SELECT statement. In the query,
GROUP BY clause is placed after the WHERE clause. In the query,
GROUP BY clause is placed before ORDER BY clause if used any.
SELECT EXTRACT(DAY FROM transaction_time) AS Jan20_day, SUM(transaction_amount),
SUM(transaction_amount)
OVER(ORDER BY EXTRACT(DAY FROM transaction_time)
ROWS BETWEEN 2 PRECEDING AND CURRENT ROW) AS rolling_average
FROM sales
WHERE EXTRACT(MONTH FROM transaction_time)=1
GROUP BY {property}
AND EXTRACT(YEAR FROM transaction_time)=2020 GROUP BY EXTRACT(DAY FROM transaction_time)
**Is there a way to count how many strings in a specific column are seen for
Since the value in the column 2 gets repeated sometimes due to the fact that some clients make several transactions in different times (the client can make a transaction in the 1st month then later in the next year).
Is there a way for me to count how many IDs are completely new per month through a group by (never seen before)?
Please let me know if you need more context.
Thanks!
A simple way is two levels of aggregation. The inner level gets the first date for each customer. The outer summarizes by year and month:
select year(min_date), month(min_date), count(*) as num_firsts
from (select customerid, min(date) as min_date
from t
group by customerid
) c
group by year(min_date), month(min_date)
order by year(min_date), month(min_date);
Note that date/time functions depends on the database you are using, so the syntax for getting the year/month from the date may differ in your database.
You can do the following which will assign a rank to each of the transactions which are unique for that particular customer_id (rank 1 therefore will mean that it is the first order for that customer_id)
The above is included in an inline view and the inline view is then queried to give you the month and the count of the customer id for that month ONLY if their rank = 1.
I have tested on Oracle and works as expected.
SELECT DISTINCT
EXTRACT(MONTH FROM date_of_transaction) AS month,
COUNT(customer_id)
FROM
(
SELECT
date_of_transaction,
customer_id,
RANK() OVER(PARTITION BY customer_id
ORDER BY
date_of_transaction ASC
) AS rank
FROM
table_1
)
WHERE
rank = 1
GROUP BY
EXTRACT(MONTH FROM date_of_transaction)
ORDER BY
EXTRACT(MONTH FROM date_of_transaction) ASC;
Firstly you should generate associate every ID with year and month which are completely new then count, while grouping by year and month:
SELECT count(*) as new_customers, extract(year from t1.date) as year,
extract(month from t1.date) as month FROM table t1
WHERE not exists (SELECT 1 FROM table t2 WHERE t1.id==t2.id AND t2.date<t1.date)
GROUP BY year, month;
Your results will contain, new customer count, year and month
database mode
The only relevant table is 'employee' in the database model.
Asked: In which month are the most employee's birthdays?
By using
SELECT DATEPART(m, dateofbirth) AS month
FROM employee
I can actually see all the months for every employee and count it myself.
But how can I show the most common birthday month?
Thanks in advance!
recent output (for comment below)
You need to use GROUP BY. This groups up the separate month values. Once you've done that, you can apply COUNT, and then order the values in descending order on that statistic. Then you need to wrap that logic in a Common Table Expression, so you can select just the months that have the maximum COUNT.
WITH ranking AS (
SELECT
DATEPART(m, dateofbirth) AS month,
COUNT(*) as ct
FROM DM_MTA.dbo.employee
GROUP BY DATEPART(m, dateofbirth)
)
select
month
from
ranking
where ct = (select max(ct) from ranking)
This would give you exact month you're looking for:
SELECT TOP 1 DATEPART(m, dateofbirth) AS month
FROM employee
GROUP BY DATEPART(m, dateofbirth)
ORDER BY count(DATEPART(m, dateofbirth)) DESC
I have db with names etc with date of birth. How can I get count of columns for all 12 months of the dates?
Exact code depends on the database you use; you should, somehow, "extract" month from date of birth in order to GROUP BY it.
In Oracle, you might have done it as
select to_char(date_of_birth), 'mon') dob_month,
count(*)
from your_table
group by to_char(date_of_birth, 'mon');
or
select extract(month from date_of_birth) dob_month,
count(*)
from your_table
group by extract(month from date_of_birth);